Fill Ups of Some Basic Concepts of Chemistry - SelfStudys

Fill Ups of Some Basic Concepts of Chemistry

Q.1. The modern atomic mass unit is based on ........................... (1980)

Ans. Carbon-12.

Sol. Carbon-12.

Q.2. The total number of electrons present in 18 ml of water is .......................

(1980)

Ans. 6.02 × 1024

Sol. 6.02 × 1024

18 ml H2O = 18 g H2O (∵ density of water = 1 g/cc)

= 1 mole of H2O.

1 Mole of H2O = 10 × 6.02 × 1023 electrons

(∵ Number of electrons present in one molecule of water

= 2 + 8 = 10)

= 6.02 × 1024 electrons

Q.3. 3 g of a salt of molecular weight 30 is dissolved in 250 g of water. The

molality of the solution is ............... . (1983 - 1 Mark)

Q.4. The weight of 1 × 1022 molecules of CuSO4.5H2O is ............... (1991 - 1 Mark)

Ans. 4.14 g

Sol. TIPS/Formulae : 1 Mole = 6.023 × 1023 molecules = Molecular weight in gms.

Weight of 6.023 × 1023 (Avogadro’s number) molecules of CuSO4.5H2O = Molecular

wt. of CuSO4.5H2O = 249 g.

∴ Weight of 1 × 1022 molecules of CuSO4.5H2O

Q.5. The compound YBa 2 Cu3 O7 , which shows super conductivity, has copper

in oxidation state..................., assume that the rare earth element yttrium is in its

usual + 3 oxidation state. (1994 - 1 Mark)

Sol. NOTE : Sum of oxidation states of all atoms (elements) in a neutral compound

is zero.

TIPS/Formulae : As YBa2Cu3O7 is neutral. (+ 3) + 2 (+ 2) + 3 (x) + 7 (–2) = 0 or 3 +

4 + 3x – 14 = 0

Integar Type ques of Some Basic Concepts of

Chemistry

Q. 1. A student performs a titration with different burettes and finds titre values

of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the

average titre value is (2010)

Q. 2. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The

number of silver atoms on a surface of area 10–12 m2 can be expressed in

scientific notation as y × 10x. The value of x is : (2010)

Q. 3. The difference in the oxidation numbers of the two types of sulphur atoms

in Na2S4O6 is (2011)

Q. 4. If the value of Avogadro number is 6.023 × 1023 mol–1 and the value of

Boltzmann constant is 1.380 × 10–23 J K–1, then the number of significant digits

in the calculated value of the universal gas constant is (JEE Adv. 2014)

Ans. Sol. R = NA × k = 6.023 × 1023 × 1.380 × 10–23

= 8.312 which has 4 significant figures

Subjective questions of Some Basic Concepts of

Chemistry (Part -1)

Q. 1. What weight of AgCl will be precipitated when a solution containing 4.77 g

of NaCl is added to a solution of 5.77 g of AgNO3?

(1978)

Ans. Sol. TIPS/Formulae : Write the balance chemical equation and use mole

concept for limiting reagent.

Q. 2. One gram of an alloy of aluminium and magnesium when treated with excess

of dil. HCl forms magnesium chloride, aluminium chloride and hydrogen. The

evolved hydrogen, collected over mercury at 0°C has a volume of 1.20 litres at 0.92

atm. pressure. Calculate the composition of the alloy. [H = 1, Mg = 24, Al = 27]

(1978)

Ans. Sol. TIPS/Formulae : (i) Find volume of H2 at N.T.P. (ii) Total amount of

H2 liberated = H2liberated by Mg & HCl + H2 liberated by Al & HCl.

Conversion of volume of H2 to N.T.P

Given conditions N.T.P conditions

P1 = 0.92 atm. P2 = 1 atm.

V1 = 1.20 litres V2 = ?

T1 = 0 + 273 = 273 K T2 = 273 K

Wt. of alloy = 1 g Let the wt. of aluminium in alloy = x g

∴ Wt. of magnesium in alloy = (1 – x) g According to equation

(i) 54 g of Al = 67200 ml of H2 at N.T.P ∴ x g of Al = 6 = 1244.4 x ml of

H2 at N.T.P

(ii) 24 g of Mg = 22400 ml of H2 at N.T.P (1 – x) g of Mg = × (1 – x) = 933.3

(1 – x) ml of H2

Hence total vol. of H2 collected at N.T.P = 1244.4 x + 933.3 (1 – x) ml

But total vol. of H2 as calculated above = 1104 ml

∴ 1244.4 x + 933.3 (1 – x) = 1104 ml

1244.4 x – 933.3 x = 1104 – 933.3

311.1 x = 170.7, x = 0.5487

Hence 1 g of alloy contains Al = 0.5487 g

Q. 3. Igniting MnO2 converts it quantitatively to Mn3O4. A sample of pyrolusite

is of the following composition : MnO2 80%, SiO2 and other inert constituents

15%, rest being water. The sample is ignited in air to constant weight. What is

the percentage of Mn in the ignited sample? (1978) [O = 16, Mn = 54.9]

Ans. Sol.

Let the amount of pyrolusite ignited = 100.00 g

∴ Wt. of MnO2 = 80 g (80% of 100 g = 80 g) Wt. of SiO2 and other inert substances =

15 g Wt. of water = 100 – (80 + 15) = 5 g

According to equation, 260.7 g of MnO2 gives = 228.7 g of Mn3O4

NOTE :

During ignition, H2O present in pyrolusite is removed while silica and other inert

substances remain as such.

∴ Total wt. of the residue = 70.2 + 15 = 85.2 g

Calculation of % of Mn in ignited Mn3O4

3 Mn = Mn3O4

3 × 54.9 = 164.7 g 3 × 54.9 + 64 = 228.7g

Since, 228.7 g of Mn3O4 contains 164.7 g of Mn

Q. 4. 4.215 g of a metallic carbonate was heated in a hard glass tube and the

CO2 evolved was found to measure 1336 ml at 27°C and 700 mm pressure. What is

the equivalent weight of the metal? (1979)

Ans. Sol. TIPS/Formulae : (i) Find the volume of CO2 at NTP

(ii) Find molecular wt. of metal carbonate

(iii) Find the wt. of metal

(iv) Calculate equivalent weight of metal

Given P1 = 700 mm, P2 = 760 mm, V1 = 1336 ml, V2 = ?

T1 = 300 K, T2 = 273 K

Q. 5. (a) 5.5 g of a mixture of FeSO4. 7H2O and Fe2(SO4)3. 9H2O requires 5.4 ml

of 0.1 N KMnO4 solution for complete oxidation. Calculate the number of gram

mole of hydrated ferric sulphate in the mixture.

(b) The vapour density (hydrogen = 1) of a mixture consisting of NO2 and

N2O4 is 38.3 at 26.7°C. Calculate the number of moles of NO2 in 100 g of the

mixture. (1979)

Ans. Sol. (a) Equivalents of KMnO4 = Equivalents of FeSO4 . 7H2O

5.4 ml 0.1 N KMnO4 = = 5.4 × 10–4 equivalentsAmount of FeSO4 = 5.4 × 10–4 ×

Mol wt. of

FeSO4.7H2O

= 5.4 × 10–4 × 278 = 0.150 g

Total weight of mixture = 5.5 g

Amount of ferric sulphate = 5.5 – 0.150 g = 5.35 g

(b) Using the relation, Mol. wt. = 2 × vapour density, we get Mol. wt. = 2 × 38.3 =

76.6

Then weight of N2O4 in mixture = 100 – x No. of moles of NO2 = ....(ii)

Q. 6. 5 ml of a gas containing only carbon and hydrogen were mixed with an

excess of oxygen (30 ml) and the mixture exploded by means of an electric spark.

After the explosion, the volume of the mixed gases remaining was 25 ml. On

adding a concentrated solution of potassium hydroxide, the volume further

diminished to 15 ml of the residual gas being pure oxygen. All volumes have

been reduced to N.T.P.

Calculate the molecular formula of the hydrocarbon gas. (1979)

Ans. Sol. Volume of oxygen taken = 30 ml, Volume of unused oxygen = 15 ml

Volume of O2 used = Volume of O2 added – Volume of O2 left

= 30 – 15 = 15 ml

Volume of CO2 produced = Volume of gaseous mixture after explosion – Volume of

unused oxygen or Volume of CO2 produced = 25 – 15 = 10 ml

Volume of hydrocarbon = 5 ml General equation for combustion of a hydrocarbon is

as follows –

Q. 7. In the analysis of 0.500 g sample of feldspar, a mixture of chlorides of sodium

and potassium is obtained which weighs 0.1180g. Subsequent treatment of mixed

chlorides with silver nitrate gives 0.2451g of silver chloride. What is the percentage

of sodium oxide and potassium oxide in feldspar.

(1979)

Ans. Sol. TIPS/Formulae : (i) Equate given mass of AgCl against mass obtained from

NaCl and KCl

(ii) 2NaCl ≡ Na2O & 2KCl ≡ K2O Let amount of NaCl in mixture = x gm

∴ amount of KCl in mixture = (0.118 - x) gm

NaCl + AgNO3 —→ AgCl + NaNO3 58.5 g

143.5 g

∵ 58.5 g NaCl gives AgCl = 143.5g

∴ x = 0.0338g

∴ Amount of NaCl in mixture = 0.0338g

∴ Amount of KCl in mixture = 0.118 – 0.0338 = 0.0842g

Since

2NaCl ≡ Na2O

2 × 58.5 62

= 117.0

Q. 8. A compound contains 28 percent of nitrogen and 72 percent of metal by

weight. 3 atoms of metal combine with 2 atoms of N. Find the atomic weight of

metal. (1980)

Ans. Sol. According to problem, three atoms of M combine with 2 atoms of N

∴ Formula of compound is M3N2 (Where M is the metal) Equivalent wt of N = 14/3 (∵

valency of N in compound is 3)

∵ 28 g N combines with = 72g metal

∴ Eq. wt. of metal = 12

At wt of metal = Eq. wt × valency = 12 × 2 = 24 [Valency of metal = 2]

Q. 9. (i) A sample of MnSO4.4H2O is strongly heated in air. The residue is

Mn3O4.

(ii) The residue is dissolved in 100 ml of 0.1 N FeSO4 containing dilute H2SO4.

(iii) The solution reacts completely with 50 ml of KMnO4 solution.

(iv) 25 ml of the KMnO4 solution used in step (iii) requires 30 ml of 0.1 N

FeSO4 solution for complete reaction.

Find the amount of MnSO4.4H2O present in the sample. (1980)

Ans. Sol. Following reactions take place

Mn3O4 + 2FeSO4 + 4H2SO4 —→ Fe2(SO4)3 + 3MnSO4 + 4H2O

Milliequivalents of FeSO4 in 30 ml of 0.1N FeSO4 = 30 × 0.1 = 3 m. eq.

According to problem step (iv) 25 ml of KMnO4 reacts with = 3 m eq of FeSO4

Thus in step (iii) of the problem, 50 ml of KMnO4 reacts with = 3/25 x m.eq. of

FeSO4

= 6 meq of FeSO4

Milli eq. of 100 ml of 0.1N FeSO4 = 100 × 0.1 = 10 m eq.

FeSO4 which reacted with Mn3O4 = (10–6) = 4 m eq.

Milli eq of FeSO4 = Milli eq. of Mn3O4 (∵ Milli eq of oxidising agent and reducing

agent are equal)

∵ Mn3O4 ≡ 3MnSO4 .4H2O

∴ 1 Meq of Mn3O4 = 3 Meq of MnSO4 . 4H2O

∴ 4 Meq of Mn3O4 = 12 Meq of MnSO4 . 4H2O

Wt of MnSO4.4H2O in sample = 12 × 111.5 = 1338 mg = 1.338g.

Q. 10. (a) One litre of a sample of hard water contains 1 mg of CaCl2 and 1 mg

of MgCl2. Find the total hardness in terms of parts of CaCO3 per 106 parts of

water by weight.

(b) A sample of hard water contains 20 mg of Ca++ ions per litre. How many

milli-equivalent of Na2CO3 would be required to soften 1 litre of the sample?

(c) 1 gm of Mg is burnt in a closed vessel which contains 0.5 gm of O2.

(i) Which reactant is left in excess?

(ii) Find the weight of the excess reactants?

(iii) How may milliliters of 0.5 N H2SO4 will dissolve the residue in the vessel.

(1980)

Ans. Sol. (a)

∴ Total CaCO3 formed by 1 mg CaCl2 and 1 mg MgCl2 = 0.90 + 1.05 = 1.95 mg

∴ Amount of CaCO3 present per litre of water = 1.95mg

∴ wt of 1 ml of water = 1g = 103 mg

∴ wt of 1000 ml of water = 103 × 103 = 106mg

∴ Total hardness of water in terms of parts of CaCO3 per 106 parts of water by weight

= 1.95 parts.

Q. 11. A hydrocarbon contains 10.5g of carbon per gram of hydrogen. 1 litre of the

vapour of the hydrocarbon at 127°C and 1 atmosphere pressure weighs 2.8g. Find

the molecular formula. (1980)

Ans. Sol.

Given P = 1 atm V = 1L, T = 127°C = 127 + 273 = 400 K

PV = nRT (Ideal gas equation)

Q. 12. Find (1980) (i) The total number of neutrons and (ii) The total mass of neutron

in 7 mg of 14C. (Assume that mass of neutron = mass of hydrogen atom)

Ans. Sol.

(i) No. of C atoms in 14g of 14C = 6.02 × 1023

∴ No. of C atom in 7 mg (7/1000g) of 14C

Q. 13. A mixture contains NaCl and unknown chloride MCl.

(i) 1 g of this is dissolved in water. Excess of acidified AgNO3 solution is added to

it. 2.567 g of white ppt. is formed.

(ii) 1 g of original mixture is heated to 300°C. Some vapours come out which are

absorbed in acidified AgNO3 solution, 1.341 g of white precipitate was obtained.

Find the molecular weight of unknown chloride. (1980)

Ans. Sol. Weight of AgCl formed = 2.567 g

Amount of AgCl formed due to MCl = 1.341 g (∵ NaCl does not decompose on heating

to 300°C)

∴ Weight of AgCl formed due to NaCl = 2.567 – 1.341= 1.226g

Q. 14. A 1.00 gm sample of H2O2 solution containing X per cent H2O2 by weight

requires X ml of a KMnO4 solution for complete oxidation under acidic conditions.

Calculate the normality of the KMnO4 solution. (1981 - 3 Marks)

Ans. Sol. The complete oxidation under acidic conditions can be represented as

follows:

5H2O2 + 2MnO4- + 6H+ → 5O2 + 2Mn2++ 8H2O

Since 34 g of H2O2 = 2000 ml of 1N . H2O2

Subjective questions of Some Basic Concepts of

Chemistry (Part -2)

Q. 15. Balance the following equations.

(i) Cu2O + H+ + NO3- → Cu2+ + NO + H2O (1981 - 1 Mark)

(ii)K4[Fe(CN)6] + H2SO4 + H2O → K2SO4 + FeSO4 + (NH4)2SO4 + CO (1981 - 1

Mark)

(iii) C2H5OH + I2 + OH– → CHI3 + HCO 3- + I– + H2O (1981 - 1 Mark)

Ans. Sol. Balance the reactions by ion electron method.

(i) Cu2O + 2H+ → 2Cu2+ + H2O + 2e–] × 3 ......(i)

(ii)K4[Fe(CN)6] + 6H2SO4 + 6H2O

→ 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO

(iii) C2H5OH + 4I2 + 8OH–

→CHI3 + HCO3- + 5I– + 6H2O

Q. 16. Hydroxylamine reduces iron (III) according to the equation:

2NH2OH + 4Fe3+ → N2O(g) ↑ + H2O + 4 Fe2+ + 4H+

Iron (II) thus produced is estimated by titration with a standard permanganate

solution. The reaction is :

MnO4- + 5 Fe2+ + 8H+ → Mn2+ + 5 Fe3+ + 4H2O

A 10 ml. sample of hydroxylamine solution was diluted to 1 litre. 50 ml. of this

diluted solution was boiled with an excess of iron (III) solution. The resulting

solution required 12 ml. of 0.02 M KMnO4 solution for complete oxidation of

iron (II). Calculate the weight of hydroxylamine in one litre of the original

solution. (H = 1, N = 14, O = 16, K = 39, Mn = 55, Fe = 56) (1982 - 4 Marks)

Ans. Sol. Given 2NH2 OH + 4Fe3+ → N2O + H2O + 4Fe2++ 4H+ .....(i) and MnO4- +

5Fe 2+ + 8H+ → Mn2+ + 5Fe3++ 4H2O ..(ii)

∴ 10 NH2OH + 4MnO4- + 12H+ → 5N2O + 21H2O+ 4Mn2+ [On multiplying (i) by 5 and

(ii) by 4 and then adding the resulting equations]

Molecular weight of NH2OH = 33

Thus 4000 ml of 1M MnO4– would react with NH2OH = 330g

∴ 12 ml of 0.02 M KMnO4 would react with NH2OH

Q. 17. The density of a 3 M sodium thiosulphate solution (Na2S2O3) is 1.25 g per

ml. Calculate (i) the percentage by weight of sodium thiosulphate, (ii) the mole

fraction of sodium thiosulphate and (iii) the molalities of Na+ and S2O32– ions. (1983

- 5 Marks)

Ans. Sol. TIPS/Formulae :

(ii) 1 mole of Na2S2O3 gives 2 moles of Na+ and 1 mole of S2O3

2– Molecular wt. of

sodium thiosulphate solution (Na2S2O3)

= 23 × 2 + 32 × 2 + 16 × 3= 158

[Wt. of Na2S2O3 = Molarity × Mol wt] (ii) Mass of 1 litre solution = 1.25 × 1000 g =

1250 g

[∵density = 1.25g/l]

Mole fraction of Na2S2O3

(iii) 1 mole of sodium thiosulphate (Na2S2O3) yields 2 moles of Na+ and 1 mole of

S2 O32- .

Q. 18. 4.08 g of a mixture of BaO and an unknown carbonate MCO3 was heated

strongly. The residue weighed 3.64 g. This was dissolved in 100 ml of 1 N HCl.

The excess acid required 16 ml of 2.5 N NaOH solution for complete

neutralization.

Identify the metal M. (1983 - 4 Marks) (At. wt. H = 1, C = 12, O = 16, Cl = 35.5,

Ba = 138)

Ans. Sol. Weight of MCO3 and BaO = 4.08 g (given)

Weight of residue = 3.64 g (given)

∴ Weight of CO2 evolved on heating = (4.08 – 3.64) g = 0.44 g

Volume of 1N HCl in which residue is dissolved = 100 ml

Volume of 1N HCl used for dissolution = (100 – 2.5 × 16) ml = 60 ml

Q. 19. Complete and balance the following reactions :

(i) Zn + NO3- → Zn2+ + NH +

4 (1983 - 1 Mark)

(ii) Cr 2 O72- + C2H4O → C2H4O2 + Cr3+ (1983 - 1 Mark)

(iii) HNO3 + HCl → NO + Cl2 (1983 - 1 Mark)

(iv) Ce3+ + S2O82- → SO2

4- + Ce4+ (1983 - 1 Mark)

(v) Cl2 + OH– → Cl– + ClO– (1983 - 1 Mark)

(vi) Mn2+ + PbO2 → MnO -4 + H2O (1986 - 1 Mark)

(vii) S + OH– → S2– + S2O3

2- (1986 - 1 Mark)

(viii) ClO 3- + I– + H2SO4 → Cl– + HSO-

4 (1986 - 1 Mark)

(ix) Ag++ AsH3 → H3AsO3 + H+ (1986 - 1 Mark)

Ans. Sol. TIPS/Formulae : Balance the atoms as well as charges by ion electron/

oxidation number method.

While balancing the equations, both the charges and atoms must balance.

(i) 4Zn + NO 3- + 10H+ —→ 4Zn2+ + NH +4 + 3H2O

(ii) Cr2O2-7 + 3C2H4O + 8H+ —→ 3C2H4O2 + 2Cr3+ + 4H2O

(iii) 2HNO3 + 6HCl —→ 2NO + 3Cl2 + 4H2O

(iv) 2Ce3+ + S2O82- —→ 2 SO4

2-+ 2Ce4+

(v) Cl2 + 2OH– —→ Cl– + ClO– + H2O

(vi) 2Mn2+ + 5PbO2 + 4H+ → 2 MnO -4 + 2H2O + 5Pb2+

(vii) 4S + 6OH– → 2S2– + S2 O32- + 3H2O

(viii) ClO 3- + 6I– + 6H2SO4 → Cl– + 6 HSO -4 + 3I2 + 3H2O

(ix) 6Ag++ AsH3 + 3H2O → 6Ag + H3AsO3 + 6H+

Q. 20. 2.68 × 10–3 moles of a solution containing an ion An+ require 1.61 × 10–

3 moles of MnO-4

for the oxidation of An+ to AO3- in acid medium. What is the value

of n? (1984 - 2 Marks)

Ans. Sol. TIPS/Formulae : Equivalents of A oxidised = Equivalents of A reduced.

Since in acidic medium, An+ is oxidised to AO3–, the change in oxidation state from

(+5) to (+n) = 5 – n [∵ O.S. of A in AO3- =+5]

∴ Total number of electrons that have been given out during oxidation of 2.68 × 10–

3 moles of An+

= 2.68 × 10–3 × (5 – n)

Thus the number of electrons added to reduce 1.61 × 10–3 moles of MnO-4 to Mn2+,

i.e. (+7) to (+2) =1.61 × 10–3 × 5

[Number of electrons involved = + 7 – (+2) = 5]

∴ 1.61 × 10–3 × 5 = 2.68 × 10–3 × (5 – n)

Q. 21. Five ml of 8N nitric acid, 4.8 ml of 5N hydrochloric acid and a certain

volume of 17M sulphuric acid are mixed together and made upto 2litre. Thirty ml.

of this acid mixture exactly neutralise 42.9 ml of sodium carbonate solution

containing one gram of Na2CO3.10H2O in 100 ml. of water. Calculate the amount

in gram of the sulphate ions in solution. (1985 - 4 Marks)

Ans. Sol. TIPS/Formulae : (i) Find normality of acid mixture and Na2CO3 . 10H2O.

Equate them to find volume of H2SO4.

(iii) Equivalent of SO4

2– = equivalents of H2SO4 × Eq. wt. of SO4– –

N × V (ml.) = meq.

Acid mixture contains 5 ml of 8N, HNO3, 4.8 ml of 5N, HCl and say, ‘V’ ml of 17 M

≡ 34 N, H2SO4. [1MH2SO4 = 2N.H2SO4]

N1V1 = N2V2

Q. 22. Arrange the following in increasing oxidation number of iodine. (1986 - 1

Mark)

I2, HI, HIO4, ICl

Ans. Sol. HI < I2 < ICl < HIO4; O.N. of I in I2 = 0, HI = –1, ICl = +1, HIO4 = +7.

Q. 23. (i) What is the weight of sodium bromate and molarity of solution

necessary to prepare 85.5 ml of 0.672 N solution when the half-cell reaction is -

BrO3- + 6H+ + 6e– → Br– + 3H2O

(ii) What would be the weight as well as molarity if the half-cell reaction is : -

2 BrO3- + 12H+ + 10e– → Br2 + 6H2O (1987 - 5 Marks)

Ans. Sol. (i) From the given half-cell reaction, Here Eq. wt. of NaBrO3

[∵ number of electron involved = 6]

Now we know that Meq. = Normality × Vol. in ml. = 85.5 × 0.672 = 57.456

Q. 24. A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C12H22O11).

Calculate : (i) molal concentration and (ii) mole fraction of sugar in the syrup.

(1988 - 2 Marks)

Ans. Sol. (i) Weight of sugar syrup = 214.2 g Weight of sugar in the syrup = 34.2 g

∴ Weight of water in the syrup = 214.2 – 34.2 = 180.0 g Mol. wt. of sugar, C12H22O11 =

342

Q. 25. A sample of hydrazine sulphate (N2H6SO4) was dissolved in 100 ml. of

water, 10 ml of this solution was reacted with excess of ferric chloride solution and

warmed to complete the reaction. Ferrous ion formed was estimated and it

required 20 ml. of M/50 potassium permanganate solution.

Estimate the amount of hydrazine sulphate in one litre of the solution. (1988 - 3

Marks)

Reaction : 4Fe3+ + N2H4 → N2 + 4Fe2+ + 4H+

MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O.

Ans. Sol. TIPS/Formulae : No. of equivalents of KMnO4 = No. of equivatents of

hydrazine sulphate.

Number of equivalents of and if weight of hydrazin

sulphate be x gm then

Q. 26. An equal volume of a reducing agent is titrated separately with 1M

KMnO4 in acid neutral and alkaline media. The volumes of KMnO4 required are

20 ml. in acid, 33.4 ml. neutral and 100 ml. in alkaline media. Find out the

oxidation state of manganese in each reduction product. Give the balanced

equations for all the three half reactions. Find out the volume of 1M

K2Cr2O7 consumed; if the same volume of the reducing agent is titrated in acid

medium.(1989 - 5 Marks)

Ans. Sol. TIPS/Formulae : No. of equivalents of KMnO4 in neutral medium = No. of

equivalents of reducing agent.

Assuming that KMnO4 shows the following changes during its oxidising nature.

Acidic medium Mn7+ + n1e– → Mna+ ∴ n1 = 7 – a

Neutral medium Mn7+ + n2e– → Mnb+ ∴ n2 = 7 – b

Alkaline medium Mn7+ + n3e– → Mnc+ ∴ n3 = 7 – c

Let V ml. of reducing agent be used for KMnO4 in different medium.

∴ Meq. of reducing agent

= Meq. of KMnO4 in acid mediumMeq. of KMnO4 in neutral medium

= Meq. of KMnO4 in alkaline medium= 1 × n1 × 20 = 1 × n2 × 33.4 = 1 × n3 × 100 =

n1 = 1.667 n2 = 5 n3

Since n1, n2 and n3 are integers and n1 is not greater than 7

∴ n3 = 1 Hence n1 = 5 and n2 = 3

∴ Different oxidation states of Mn in Acidic medium Mn7+ + 5e– → Mna+ or a = + 2

Neutral medium Mn7+ + 3e– → Mnb+ or b = + 4

Alkaline medium Mn7+ + 1e– → Mnc+ or c = + 6

Further, same volume of reducing agent is treated with K2Cr2O7, and therefore Meq. of

reducing agent = Meq. of K2Cr2O7

1 × 5 × 20= 1 × 6 × V [Q Cr+6 + 6e– → 2Cr+3]

V = 16.66 mL ∴ 1M = 6 × 1N

Q. 27. A mixture of H2C2O4 (oxalic acid) and NaHC2O4 weighing 2.02 g was

dissolved in water and solution made upto one litre. Ten millilitres of the solution

required 3.0 ml. of 0.1 N sodium hydroxide solution for complete neutralization. In

another experiment, 10.0 ml. of the same solution, in hot dilute sulphuric acid

medium. require 4.0 ml. of 0.1 N potassium permanganate solution for complete

reaction.

Calculate the amount of H2C2O4 and NaHC2O4 in the mixture. (1990 - 5 Marks)

Ans. Sol. TIPS/Formulae : No. of equivalents of KMnO4

= No. of equivatents of reducing agents.

Case I. Reaction of NaOH with H2C2O4 and NaHC2O4.

(i) H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

(ii) NaHC2O4 + NaOH → Na2C2O4 + H2O

Number of milliequivalents of NaOH = N × V = 3.0 × 0.1 = 0.3

∴ Combined normality of the mixture titrated with NaOH

Case II. Reaction of C2O4– ion and KMnO4

(iii) 5C2O4– + MnO4

– + 16H+ → 2Mn2+ + 10CO2 + 8H2O KMnO4 will react in same

manner with both NaHC2O4 and H2C2O4 as it can be seen from the above reaction.

Number of milliequivalents of KMnO4 = 4.0 × 0.1 = 0.4

∴ Combined normality of the mixture titrated with KMnO4

The difference (0.04 N – 0.03 N = 0.01 N) is due to NaHC2O4 The total normality of

NaHC2O4 will be = 0.01 + 0.01 = 0.02 N From equation (ii) in

case I.

Eq. wt. of NaHC2O4 = 112 Amount of NaHC2O4 in one litre of solution formed =

0.01 × 112 = 1.12 g and amount of H2C2O4 = 2.02 – Wt. of NaHC2O4 = 2.02 – 1.12 =

0.90 g

Q. 28. A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was

heated below 600ºC until the weight of the residue was constant. If the loss in

weight is 28.0 per cent, find the amount of lead nitrate and sodium nitrate in the

mixture. (1990 - 4 Marks)

Ans. Sol. TIPS/Formulae : Let the amount of NaNO3 in the mixture = x g

∴ The amount of Pb(NO3)2 in the mixture = (5 – x) g

Heating effect of sodium nitrate and lead nitrate

Subjective questions of Some Basic Concepts of

Chemistry (Part -3)

Q. 29. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume).

The density of the solution is 1.84 g/ml. (1990 - 1 Marks)

Ans. Sol. TIPS/Formulae :

Q. 30. A solution of 0.2 g of a compound containing Cu2+ and C22- O4 ions on

titration with 0.02 M KMnO4 in presence of H2SO4 consumes 22.6 ml. of the

oxidant. The resultant solution is neutralized with Na2CO3, acidified with dil.

acetic acid and treated with excess KI. The liberated iodine requires 11.3 ml of

0.05 M Na2S2O3 solution for complete reduction.

Find out the molar ratio of Cu2+ to C2O42- in the compound.

Write down the balanced redox reactions involved in the above titrations. (1991

- 5 Marks)

Ans. Sol. In th e given problem, a solution contain in g Cu2+ and C2O4

2- is titrated first

with KMnO4and then with Na2S2O3 in presence of KI. In titration with KMnO4, it is the

C2O24- ions that react with the MnO-

4 ions. The concerned balanced equation may be

written as given below.

2MnO4- + 5 C2O2

4- + 16H+ → 2Mn2+ + 10CO2 + 8H2O

Thus according to the above reaction 2 mmol MnO-4 ≡ 5 mmol C2 O2

4-

However, No. of mmol of MnO-4 used in titration = Vol. in ml × M = 22.6 × 0.02 =

0.452 mmol MnO-4

Since 2 mmol MnO-4 ≡ 5 mmol C2O2

4-0.452 mmol MnO-4 ≡ 5/2 × 0.452 = 1.130 mmol

C2 O24-

Balanced equations in two cases Case I. Mn+7 + 5e– → Mn+2

C2+3 → 2C+4 + 2e–

Case II. 2Cu+2 + 2e– → Cu2+ 2I– → I2 + 2e– and I2 + 2e– → 2I– 2S2+2 → S4

+3/2 + 2e–

Q. 31. A 1.0 g sample of Fe2O3 solid of 55.2% purity is dissolved in acid and

reduced by heating the solution with zinc dust.

The resultant solution is cooled and made upto 100.0 ml. An aliquot of 25.0 ml of

this solution requires 17.0 ml of 0.0167 M solution of an oxidant for titration.

Calculate the number of electrons taken up by the oxidant in the reaction of the

above titration. (1991 - 4 Marks)

Number of moles of Fe3+ ions = 2 × 3.454 × 10–3

= 6.9 × 10–3 mol = 6.90 mmolSince its only 1 electron is exchanged in the conversion

of Fe3+ to Fe2+, the molecular mass is the same as equivalent mass.

∴ Amount of Fe2+ ion in 100 ml. of sol. = 6.90 meq Volume of oxidant used for 100

ml of Fe2+ sol.

= 17 × 4 = 68 ml.

Amount of oxidant used = 68 × 0.0167 mmol

= 1.1356 mmolLet the number of electrons taken by the oxidant = n

∴ No. of meq. of oxidant used = 1.1356 × n

Q. 32. A 2.0 g sample of a mixture containing sodium carbonate, sodium

bicarbonate and sodium sulphate is gently heated till the evolution of

CO2 ceases. The volume of CO2 at 750 mm Hg pressure and at 298 K is measured

to be 123.9 ml. A 1.5g of the same sample requires 150 ml. of (M/10) HCl for

complete neutralisation. Calculate the % composition of the components of the

mixture. (1992 - 5 Marks)

On heating, the sample, only NaHCO3 undergoes decomposition as given below.

∴ % of Na2SO4 in the mixture = 100 – (42 + 26.5) = 31.5%

Q. 33. One gram of commercial AgNO3 is dissolved in 50 ml. of water. It is treated

with 50 ml. of a KI solution. The silver iodide thus precipitated is filtered off.

Excess of KI in the filterate is titrated with (M/10) KIO3 solution in presence of 6M

HCl till all I– ions are converted into ICl. It requires 50 ml. of (M/10)

KIO3 solution. 20 ml. of the same stock solution of KI requires 30 ml. of

(M/10)KIO3 under similar conditions.

Calculate the percentage of AgNO3 in the sample.

(Reaction : KIO3 + 2KI + 6HCl → 3ICl + 3KCl + 3H2O) (1992 - 4 Marks)

Ans. Sol. Reaction involved titration is 1 mole 2 moles KIO3 + 2KI + 6HCl → 3ICl +

3KCl + 3H2O

Q. 34. Upon mixing 45.0 ml. of 0.25 M lead nitrate solution with 25.0 ml of 0.10

M chromic sulphate solution, precipitation of lead sulphate takes place. How

many moles of lead sulphate are formed? Also, calculate the molar

concentrations of the species left behind in the final solution. Assume that lead

sulphate is completely insoluble. (1993 - 3 Marks)

Ans. Sol. Calculation of number of moles in 45 ml. of 0.025 M Pb(NO3)2

∴ Initial moles of Pb2+ = 0.01125

Moles of NO3- = 0.01125 × 2 = 0.02250 [1 mole Pb(NO3)2 ≡ 2 moles of NO3]

Calculation of number of moles in 25 ml. of 0.1 M chromic sulphate

Moles of chromic sulphate (Cr2(SO4)3

Moles of SO24- = 0.0025 × 3 = 0.0075 [1 Mole of chromic sulphate ≡ 3 moles of SO4

2–]

Moles of PbSO4 formed = 0.0075 [SO42– is totally consumed]

Moles of Pb2+ left = 0.01125 – 0.0075 = 0.00375

Moles of NO3- left = 0.02250 [NO3

– remain unreacted]

Moles of chromium ions = 0.0025 × 2 = 0.005

Total volume of the solution = 45 + 25 = 70 ml.

∴ Molar concentration of the species left

Q. 35. The composition of a sample of Wustite is Fe0.93O1.00 .

What percentage of the iron is present in the form of Fe (III)? (1994 - 2 Marks)

Ans. Sol. In pure iron oxide (FeO), iron and oxygen are present in the ratio 1 : 1.

However, here number of Fe2+ present = 0.93 or No. of Fe2+ ions missing = 0.07 Since

each Fe2+ ion has 2 positive charge, the total number of charge due to missing (0.07)

Fe2+ ions = 0.07 × 2 = 0.14 To maintain electrical neutrality, 0.14 positive charge is

compensated by the presence of Fe3+ ions. Now since, replacement of one Fe2+ ion by

one Fe3+ ion increases one positive charge, 0.14 positive charge must be compensated

by the presence of 0.14 Fe3+ ions.

In short, 0.93 Fe2+ ions have 0.14 Fe3+ ions 100 Fe2+ ions have = × 100 = 15.05%

Q. 36. 8.0575 × 10–2 kg of Glauber ’s salt is dissolved in water to obtain 1 dm3 of

a solution of density 1077.2 kg m–3. Calculate the molarity, molality and mole

fraction of Na2SO4 in the solution. (1994 - 3 Marks)

Ans. Sol. The formula of Glauber’s salt is Na2SO4.10H2O Molecular mass of

Na2SO4.10H2O

= [2 × 23 + 32.1 + 4 × 16] + 10 (1.01 × 2 + 16) = 322.3 g mol–1

Weight of the Glauber’s salt taken = 80.575 gm Out of 80.575 g of salt, weight of

anhydrous Na2SO4

Mole fraction of Na2SO4

Q. 37. A 3.00 g sample containing Fe3O4, Fe2O3 and an inert impure substance, is

treated with excess of KI solution in presence of dilute H2SO4.The entire iron is

converted into Fe2+ along with the liberation of iodine. The resulting solution is

diluted to 100 ml. A 20 ml of the diluted solution requires 11.0 ml of 0.5 M

Na2S2O3 solution to reduce the iodine present. A 50 ml of the diluted solution,

after complete extraction of the iodine requires 12.80 ml of 0.25 M

KMnO4 solution in dilute H2SO4 medium for the oxidation of Fe2+. Calculate the

percentages of Fe2O3 and Fe3O4 in the original sample. (1996 - 5 Marks)

Ans. Sol. TIPS/Formulae : Find the milliequivalents and equate them as per data given

in question.

For Fe3O4 → 3FeO

2e + Fe3(8/3)+ → 3Fe2+

Thus, valence factor for Fe3O4 is 2 and for FeO is 2/3.

For, Fe2O3 → 2FeO; 2e + Fe23+ → 2Fe2+ ...(1)

Thus valence factor for Fe2O3 is 2 and for FeO is 1.

Let Meq.of Fe3O4 and Fe2O3 be a and b respectively.

∴ Meq. of Fe3O4 + Meq. Fe2O3 = Meq. of I2 liberated

= Meq. of hypo used

Now, the Fe2+ ions are again oxidised to Fe3+ by KMnO4.

Note that in the change Fe2+ → Fe3+ + e–; valence factor of Fe2+ is l.

Thus, Meq. of Fe2+ (from Fe3O4) + Meq. of Fe2+ (from Fe2O3) = Meq. of

KMnO4 used .... (2)

If valence factor for Fe2+ is 2/3 from Eq. (1), then Meq. of Fe2+ (from Fe3O4) = a

If valence factor for Fe2+ is 1 then Meq. of Fe2+ (from Fe3O4) = 3a/2 … (3)

Similarly, from Eq. (2), Meq. of Fe2+ from (Fe2O3) = b.

∴ 3a/2 + b = 0.25 x 5 x 12.8 x 100/50 = 32 or 3a + 2b = 64 ....(4)

From Eqs. (3) and (4)

Meq. of Fe3O4 = a = 9 & Meq. of Fe2O3 = b = 18.5

Q. 38. An aqueous solution containing 0.10 g KIO3 (formula weight = 214.0) was

treated with an excess of KI solution. The solution was acidified with HCl. The

liberated I2consumed 45.0 mL of thiosulphate solution to decolourise the blue

starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

(1998 - 5 Marks)

Ans. Sol. TIPS/Formulae : Write the reactions taking place, balance them and equate

moles of I2 and Na2S2O3.

KIO3 + 5KI → 3K2O + 3I2 i.e., 2 I5+ + 10e– → I02

2I– → I02 + 2e–

Now liberated I2 reacts with Na2S2O3 I2 + 2e– → 2I–

2S2O32- → S4O6

2- + 2e–

∴ millimole ratio of I2 : S2O3 = 1 : 2

Q. 39. How many millilitres of 0.5 M H2SO4 are needed to dissolve 0.5 g of

copper(II) carbonate? (1999 - 3 Marks)

Ans. Sol. TIPS/Formulae : Use molarity equation to find volume of H2SO4 solution.

Q. 40. A plant virus is found to consist of uniform cylindrical particles of 150 Å

in diameter and 5000 Å long. The specific volume of the virus is 0.75 cm3/g. If

the virus is considered to be a single particle, find its molar mass.(1999 - 3

Marks)

Ans. Sol. TIPS/Formulae : (i) Volume of virus = πr2ℓ (Volume of cylinder)

Q. 41. Hydrogen peroxide solution (20 ml) reacts quantitatively with a solution

of KMnO4(20 ml) acidified with dilute H2SO4.

The same volume of the KMnO4 solution is just decolourised by 10 ml of

MnSO4 in neutral medium simultaneously forming a dark brown precipitate of

hydrated MnO2. The brown precipitate is dissolved in 10 ml of 0.2 M sodium

oxalate under boiling condition in the presence of dilute H2SO4. Write the

balanced equations involved in the reactions and calculate the molarity of H2O2.

(2001 - 5 Marks)

Ans. Sol. TIPS/Formulae : Write the balanced chemical reaction for change and apply

mole concept.

The given reactions are MnO2 ↓ + Na 2C2 O4+ 2H2SO4

—→ MnSO4 + CO2 + Na 2SO4+ 2H2O

∴ Meq. of MnO2 ≡ Meq of Na2C2O4 = 10 × 0.2 × 2 = 4

Now 2KMnO4 + 3MnSO4+ 2H2O

—→ 5MnO2 ↓ + K2SO4+ 2H2O

Since eq. wt. of MnO2 is derived from KMnO4 and MnSO4 both, thus it is better to

proceed by mole concept

mM of KMnO4 ≡ mM of MnO2 × (2/5) = 4/5

Also 5H2O2 + 2KMnO4+ 3H2SO4

—→ 2MnSO4 + K2SO4 + 8H2O+ 5O2

2KMnO4 + 5H2O2 + 3H2SO4 —→ K2SO4 + 2MnSO4 + 8H2O + 5O2

2KMnO4 + 3MnSO4 + 2H2O —→ 5MnO2 + 2H2SO4 + K2SO4

MnO2 + Na2C2O4 + 2H2SO4 —→MnSO4 + 2CO2 + Na2SO4 + 2H2O

Q. 42. Calculate the molarity of water if its density is 1000 kg/m3. (2003 - 2 Marks)

Ans. Sol. 1 litre water = 1 kg i.e. 1000 g water (∵ d = 1000 kg/m3)

So, molarity of water = 55.55M